Tentuka persamaan garis yang tegak lurus garis 2x-y+3 =0 dan melalui titik (5,3) !

PERSAMAAN GARIS LURUS

• 2x – y + 3 = 0
-y = -2x – 3
y = 2x + 3

• gradien garis y = 2x + 3
gradien ( m1) = 2

• Tegak lurus
m1 x m2 = -1
2 x m2 = -1
m2 = -1/2

• garis yg melalui titik (5,3)
y = m (x – x1) + y1
y = -1/2 (x – 5) + 3
y = – 1/2x + 5/2 + 3
y = -1/2x + 3 5/2
y = -1/2x + 5 1/2
y = -1/2x + 11/2

semoga membantu 🙂

Diket :
2x-y+3 = 0
x1 = 5
y1 = 3

Jawab :
2x-y+3 = 0
2x+3 = y
m1 = 2

m2 = -1/m1 = -1/2

y-y1 = m(x-x1)
y-3 = -1/2(x-5)
y = -1/2x+5/2+6/2
y = -1/2x+11/2

Maaf jika ada kesalahan 🙂